Question

textbook videos + if ( f(x)=\frac{g(x)}{h(x)} ), then ( f^{prime}(1)= ) question help: video message instructor submit question jump to answer

Explanation:

Step1: Find the slope of \(g(x)\)

The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). For \(g(x)\), using the points \((0,9)\) and \((4,1)\), \(m_{g}=\frac{1 - 9}{4-0}=\frac{-8}{4}=-2\). So \(g^{\prime}(x)=-2\) (since \(g(x)\) is a linear function \(y = mx + b\) and its derivative is the slope).

Step2: Find the slope of \(h(x)\) for \(x < 3\)

For \(h(x)\) (left - hand side of the vertex at \(x = 3\)), using the points \((0,2.5)\) and \((3,5)\). The slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\), so \(m_{h}=\frac{5 - 2.5}{3-0}=\frac{2.5}{3}=\frac{5}{6}\). So \(h^{\prime}(x)=\frac{5}{6}\) for \(x<3\).

Step3: Use the quotient rule

The quotient rule states that if \(f(x)=\frac{g(x)}{h(x)}\), then \(f^{\prime}(x)=\frac{g^{\prime}(x)h(x)-g(x)h^{\prime}(x)}{[h(x)]^{2}}\).
First, find \(g(1)\) and \(h(1)\). For \(g(x)\), \(g(x)=-2x + 9\) (using \(y=mx + b\), \(m=-2\), \(b = 9\)), so \(g(1)=-2(1)+9 = 7\). For \(h(x)\), \(h(x)=\frac{5}{6}x+2.5\) (using \(y=mx + b\), \(m = \frac{5}{6}\), \(b = 2.5\)), so \(h(1)=\frac{5}{6}(1)+2.5=\frac{5 + 15}{6}=\frac{20}{6}=\frac{10}{3}\).
Now, substitute into the quotient rule: \(f^{\prime}(1)=\frac{g^{\prime}(1)h(1)-g(1)h^{\prime}(1)}{[h(1)]^{2}}\). Since \(g^{\prime}(x)=-2\) (constant) and \(h^{\prime}(1)=\frac{5}{6}\) (because \(1<3\)).
\(f^{\prime}(1)=\frac{-2\times\frac{10}{3}-7\times\frac{5}{6}}{(\frac{10}{3})^{2}}\)

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Answer:

\(-\frac{9}{8}\)